For the reaction -160- 2N-2O-5 - 4NO-2 - O-2- if - - d-N-2O-5-dt - k-1-N-2O-5- -160- d-NO-2-dt - k-2-N-2O-5- -160- d-O-2-dt - k-3-N-2O-5-What is the relation between - k-1- k-2- and - k-3- k-1 - k-2 - k-3- 2k-1 - k-2 - 4k-3- 2k-1 - 4k-2 - k-3-None of the above

The correct option is B 2k_1 = k_2 = 4k_3 As we know, nbsp;for a reaction: nbsp; 2N_2O_5 → 4NO_2 + O_2 1/2 d[N_2O_5]/dt = 1/4 d[NO_ ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

For the react...

Question

For the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , if $- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]$ , $\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]$ , $\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]$ .
What is the relation between $k_{1}, k_{2}$ and $k_{3}$ ?

k_{1} = k_{2} = k_{3}

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2 k_{1} = k_{2} = 4 k_{3}

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2 k_{1} = 4 k_{2} = k_{3}

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None of the above

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Solution

The correct option is B $2 k_{1} = k_{2} = 4 k_{3}$
As we know, for a reaction: $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$
$- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = \frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{d [O_{2}]}{d t}$
So, $2 k_{1} = k_{2} = 4 k_{3} .$

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Similar questions

Q. For a reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

Given:

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation between

k_{1}, k_{2}

and

k_{3}

are:

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

What is the relation between

k_{1}, k_{2} a n d k_{3}

For the reaction;

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

Given:

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}], \frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}], \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

.
The relation between

K_{1}, K_{2}

and

K_{3}

is:

Q. For a reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

Given:

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation between

k_{1}, k_{2}

and

k_{3}

are:

Q. For the reaction

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

- \frac{Δ [N_{2} O_{5}]}{Δ t} = k_{1} [N_{2} O_{5}], + \frac{[N O_{2}]}{Δ t} = k_{2} [N_{2} O_{5}], + \frac{Δ [O_{2}]}{Δ t} = k_{3} [N_{2} O_{5}]

Thus,

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For the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , if $- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]$ , $\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]$ , $\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]$ .
What is the relation between $k_{1}, k_{2}$ and $k_{3}$ ?

The correct option is B $2 k_{1} = k_{2} = 4 k_{3}$
As we know, for a reaction: $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$
$- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = \frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{d [O_{2}]}{d t}$
So, $2 k_{1} = k_{2} = 4 k_{3} .$

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For the reaction 2N2O5→4NO2+O2, if −d[N2O5]dt=k1[N2O5], d[NO2]dt=k2[N2O5], d[O2]dt=k3[N2O5].What is the relation between k1,k2 and k3?

The correct option is B 2k1=k2=4k3As we know, for a reaction: 2N2O5→4NO2+O2−12d[N2O5]dt=14d[NO2]dt=d[O2]dtSo, 2k1=k2=4k3.

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For the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , if $- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]$ , $\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]$ , $\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]$ .
What is the relation between $k_{1}, k_{2}$ and $k_{3}$ ?

The correct option is B $2 k_{1} = k_{2} = 4 k_{3}$
As we know, for a reaction: $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$
$- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = \frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{d [O_{2}]}{d t}$
So, $2 k_{1} = k_{2} = 4 k_{3} .$